Giải thích các bước giải:

Ta có:

$3AM=MB\to \dfrac{AM}1=\dfrac{MB}3=\dfrac{AM+MB}{1+3}=\dfrac{AB}4$

$\to \dfrac{AM}{AB}=\dfrac14$

Ta có:

$NA=4NC\to \dfrac{NA}4=\dfrac{NC}1=\dfrac{NA+NC}{4+1}=\dfrac{AC}5$

$\to \dfrac{AN}{AC}=\dfrac45$

Ta có:

$\vec{BN}.\vec{CM}$

$=(\vec{AN}-\vec{AB})(\vec{AM}-\vec{AC})$

$=(\dfrac45\vec{AC}-\vec{AB})(\dfrac14\vec{AB}-\vec{AC})$

$=-\dfrac14AB^2-\dfrac45AC^2+(\dfrac45\cdot\dfrac14+1)\vec{AB}\cdot \vec{AC}$

$=-\dfrac14AB^2-\dfrac45AC^2+\dfrac65\vec{AB}\cdot \vec{AC}$

$=-\dfrac14\cdot 3^2-\dfrac45\cdot 3^2+\dfrac65\cdot 3\cdot 3\cdot \cos60^o$

$=-\dfrac{81}{20}$